Find the equation of the circle passing through the points (1,2), (3,−4), and (5,−6).
(JEE Main 2020, similar pattern)
Solution — Step by Step
The general equation is x2+y2+2gx+2fy+c=0, where the centre is (−g,−f) and radius is g2+f2−c.
We need to find g, f, and c using the three given points.
Point (1,2): 1+4+2g+4f+c=0⇒2g+4f+c=−5 … (i)
Point (3,−4): 9+16+6g−8f+c=0⇒6g−8f+c=−25 … (ii)
Point (5,−6): 25+36+10g−12f+c=0⇒10g−12f+c=−61 … (iii)
Subtract (i) from (ii): 4g−12f=−20⇒g−3f=−5 … (iv)
Subtract (ii) from (iii): 4g−4f=−36⇒g−f=−9 … (v)
Subtract (iv) from (v): 2f=−4⇒f=−2
From (v): g=−9+f=−9−2=−11
From (i): c=−5−2(−11)−4(−2)=−5+22+8=25
x2+y2−22x−4y+25=0
Centre: (11,2), Radius: 121+4−25=100=10.
Why This Works
Three non-collinear points uniquely determine a circle. The general equation has three unknowns (g, f, c), and substituting three points gives three linear equations — a solvable system.
The condition for the three points to be non-collinear is essential. If the points are collinear, the “circle” would have infinite radius (a straight line), and the system of equations would be inconsistent.
Alternative Method — Determinant form
The circle through (x1,y1), (x2,y2), (x3,y3) can be written as:
This is elegant but the 4x4 determinant expansion is time-consuming by hand. For JEE MCQs, the simultaneous equations method is usually faster.
In JEE, if the question gives three points on a circle and asks for the centre or radius, solve for g and f first (you don’t even need c if they only ask for the centre). This can save a minute of calculation.
Common Mistake
A careless error that costs marks: writing the general equation as x2+y2+gx+fy+c=0 (missing the factor of 2 in front of g and f). The standard form has 2g and 2f, so the centre is (−g,−f). If you use gx+fy instead of 2gx+2fy, your centre formula changes to (−g/2,−f/2) and things get messy. Stick to one convention consistently.
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